Rule of Association/Disjunction/Formulation 2/Proof 2

Theorem

$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$

Proof

This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.

By the tableau method:

$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$

Line		Pool	Formula	Rule	Depends upon	Notes
1			$\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$	Rule of Association: Forward Implication
2			$\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$	Rule of Association: Reverse Implication
3			$\paren {\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r} } \land \paren {\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} } }$	Rule $\text {RST} 4$	1, 2
4			$\paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$	Rule $\text {RST} 2 (3)$	3

$\blacksquare$

Sources

• 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.7$: The Derivation of Formulae: $D \, 13$